Question

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Answer 1

 

$\displaystyle\bf\\Formule~folosite:\\\\\log_aa^b=b\\\\\frac{1}{a^b}=a^{-b}\\\\\sqrt{a}=a^{\frac{1}{2}}\\\\ Rezolvare:\\\\log_2\sqrt{2}+log_3\frac{1}{\sqrt{3}}=\\\\=log_22^\frac{1}{2} +log_33^{-\frac{1}{2}}=\\\\=\frac{1}{2}+\left(-\frac{1}{2}\right)=\\\\=\frac{1}{2}-\frac{1}{2}=\boxed{\bf0}$